(a) Distance of the point (0, 0, 1) from the plane x + y + z = 3.
(a) 1/√3 (b) 2/√3 (c) √3 (d) √3/2 (Score : 1)
(b) Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to x – y + z = 0. (Scores :

Question

(a) Distance of the point (0, 0, 1) from the plane x + y + z = 3.
(a) 1/√3 (b) 2/√3 (c) √3 (d) √3/2 (Score : 1)
(b) Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to x – y + z = 0. (Scores : 3)

Accepted Answer

>

> [!TLDR]
> (a) uses the point-to-plane distance formula; (b) builds the family of planes through the line of intersection of two given planes, then fixes the parameter using the perpendicularity condition.
>
> **(a) Distance of $(0,0,1)$ from the plane $x+y+z=3$.**
>
> $$d = \frac{|0+0+1-3|}{\sqrt{1^2+1^2+1^2}} = \frac{2}{\sqrt3}$$
>
> — option (b).
>
> **(b) Plane through the line of intersection of $x+y+z=1$ and $2x+3y+4z=5$, perpendicular to $x-y+z=0$.**
>
> Family of planes through the line of intersection:
> $$(x+y+z-1) + k(2x+3y+4z-5) = 0$$
> $$(1+2k)x + (1+3k)y + (1+4k)z - (1+5k) = 0$$
>
> Its normal is $(1+2k,\ 1+3k,\ 1+4k)$. For this plane to be perpendicular to $x-y+z=0$ (normal $(1,-1,1)$), the normals must satisfy a zero dot product:
> $$(1+2k)(1) + (1+3k)(-1) + (1+4k)(1) = 0$$
> $$1+2k-1-3k+1+4k = 0 \;\Rightarrow\; 1+3k = 0 \;\Rightarrow\; k=-\frac13$$
> …

Q.(a) Distance of the point (0, 0, 1) from the plane x + y + z = 3.

Concept understanding — Equation of a Plane

Equation of a Plane

Point + normal form

Normal (perpendicular) form

Intercept form …

Q.(a) Distance of the point (0, 0, 1) from the plane x + y + z = 3.

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